Question

The velocity of a particle is given by v=12+3(t+7t²) what is the acceleration of the particle?
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Answer 1

Answer:

Explanation:

First, let us find whether the particle reverses its direction of motion during the specified time.

S=6+12t−2t2

v=dsdt=12–4t … … … (1)

Equating the velocity v with zero ( 12–4t=0 ) yields t=3s . This means particle travels for 3 seconds in forward direction and for the next 2 second in backward direction. In order to find the distance in 5 second, we have to find the magnitude of displacement in first 3 second and then 3 second to 5 second and then add the two.

Displacement in first 3s :

s(t=3s)−s(t=0)=(6+12×3−2×9)−6=18m (assuming SI Unit)

Displacement during 3 to 5 second:

s(t=5s)−s(t=3s)=(6+12×5–2×25)−(6+12×3−2×9)=−8m

Ignoring the negative sign and adding the two displacements we have the total distance 18+8=26m .

Alternatively, we can find initial velocity u by putting t = 0 in equation (1). This comes out to be 12 m/s. Thus u = 12 m/s.

Displacement in first 3 s:

s=ut+12at2=12×3+12(−4)×9=18m

Displacement in next 2s (notice that particle starts its backward journey with u=0 ):

s=ut+12at2=0−12(−4)×4=−8m